# Sum of Cubes

How to solve the given summation by using the sum of cubes (k3) formula: formula, 1 example, and its solution.

## Formula

### Formula

The sum of k3, as k goes from 1 to n, is
[n(n + 1)/2]2.

This means
13 + 23 + 33 + ... + n3 = [n(n + 1)/2]2.

## Example

### Solution

k(k2 + n) = (k3 + nk)

k3 is in the sigma.

See ∑ (+nk).
k is the variable of the sigma.
So think n as a coefficient.
Then n gets out from the sigma.
And k is in the sigma.

Summation: How to Solve

The sum of k3 is
[n(n + 1)/2]2.

Write the 'coefficient' +n.

The sum of k is
n(n + 1)/2.

Sum of k

So [n(n + 1)/2]2 + n⋅n(n + 1)/2.

[n(n + 1)/2]2
= n2(n + 1)2/22
= n2(n + 1)2/4

Power of a Product

Power of a Quotient

+n⋅n(n + 1)/2 = +n2(n + 1)/2

To combine these two fractions,
change +n2(n + 1)/2 to +2n2(n + 1)/4.

The common factor of
n2(n + 1)2/4 and +2n2(n + 1)/4
is n2(n + 1)/4.

So write
n2(n + 1)/4.

Write, n2(n + 1)2/4 ÷ n2(n + 1)/4,
[(n + 1).

And write, +2n2(n + 1)/4 ÷ n2(n + 1)/4,
+2].

[(n + 1) + 2] = (n + 3)

So n(n + 1)(n + 3)/4 is the answer.